Two definitions of $\limsup$

Question

Here are two equivalent definitions of $\limsup_{n\rightarrow\infty} a_n$:

• Let $u_n=\sup\{a_n, a_{n+1}, a_{n+2},\ldots\}$. Then $$\limsup_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty} u_n = \lim_{n\rightarrow\infty}\left(\sup\{a_n,a_{n+1},\ldots\}\right)$$

• Let $E$ be the set of all subsequential limits of $\{a_n\}$. Then $$\limsup_{n\rightarrow\infty} a_n = \sup E$$

I'm curious as to which one people usually learn first, or which one people find more intuitive. Baby Rudin only has the second definition, but doesn't mention the first as far as I know.

As far as intuition, I think of the second definition as "the biggest limit point." I'm not sure how to think of the first definition though. I can see that $u_n \ge u_{n+1}$ for all $n$, but I'm not sure how to interpret the limit of those $u_n$ conceptually.

Are there any other ways to conceptualize $\limsup$?

Answer 1

The first definition has the great virtue of exactly matching the notation: it defines $\limsup_na_n$ to be the limit (as $n\to\infty$) of the suprema (of the tails of the sequence). Since the behavior of a sequence is determined by its tails, this is a very natural thing to consider. Loosely speaking, it’s what the supremum of the sequence ‘ought’ to be if we could ignore the more or less meaningless fluctuations in every finite initial segment. We can’t quite do that literally, because there can be such fluctuations arbitrarily far out in the sequence, but we can do it in the limit. Let $u=\limsup_na_n$; any given tail of the sequence may have supremum larger than $u$, but if it does, a later tail will have a smaller supremum, having squeezed out more of the meaningless ‘early’ fluctuation. Note that because the suprema of the tails form a non-increasing sequence,

$$\limsup_na_n=\lim_{n\to\infty}\sup_{k\ge n}a_k=\inf_{n\in\Bbb N}\sup_{k\ge n}a_k\;.\tag{1}$$

This definition also generalizes relatively easily to sequences in arbitrary complete lattices, which have notions of supremum and infimum of arbitrary sets of elements. In particular, if $X$ is a set, $\wp(X)$ is a complete lattice with $\bigcup$ as supremum and $\bigcap$ as infimum. Let $\langle A_n:n\in\Bbb N\rangle$ be a sequence of subsets of $X$. A first attempt to generalize the first definition might be

$$\limsup_nA_n=\lim_{n\to\infty}\bigcup_{k\ge n}A_k\;,$$

but we don’t (yet) have a notion of the limit of a sequence of sets. The last expression in $(2)$, however, does the trick nicely: we can meaningfully define

$$\limsup_nA_n=\inf_{n\in\Bbb N}\bigcup_{k\ge n}A_k=\bigcap_{n\in\Bbb N}\bigcup_{k\ge n}A_k\;.$$

Better yet, we can see that it has the same general effect of getting rid of essentially meaningless initial fluctuations: the union of any given tail may be bigger than $\limsup_nA_n$, but if it is, a later tail will have a smaller union, having squeezed out more of the points that are in only finitely many of the $A_n$.

The second definition expresses a very important property of the limit superior of a sequence of real numbers, but I think that the first gives easier access to the various more general notions of limit superior.

Answer 2

I think a more proper definition of $\limsup$ is

$$ \limsup_{n\to\infty} a_n = \inf_{n \ge 1} \sup\{a_n, a_{n+1}, \ldots\} $$

because you need $\limsup$ and $\liminf$ to define $\lim$.

The way I think of $\limsup$ is the limit of the "upper envelope" of the sequence. Like when you have $a_n = \left(1 + \frac 1n\right)\sin n$, $\limsup_{n \to \infty} a_n = 1$ and $\liminf_{n \to \infty} a_n = -1$, but the limit doesn't exist.

Answer 3

I vaguely thought that Rudin did say explicitly that the limsup of a sequence is the supremum of its limit points. Could be wrong.

As for the first definition, you should think of the $u_n$ as being "the biggest element this side of the river". If you keep moving the river, you're left with only the biggest element. Of course this is very imprecise, but it helps me.